Introduction

Combination is the way of picking a different unique smaller set from a bigger set, without regard to the ordering (positions) of the elements (in the smaller set). This article teaches you how to find combinations. First, I will show you the technique to find combinations. Next, I will go on to explain how to use my source code. The source includes a recursive template version and a non-recursive template version. At the end of the article, I will show you how to find permutations of a smaller set from a bigger set, using both next_combination() and next_permutation().

Before all these, let me first introduce to you the technique of finding combinations.

The Technique

The notations used in this article

n: n is the larger sequence from which the r sequence is picked.

r: r is the smaller sequence picked from the n sequence.

c: c is the formula for the total number of possible combinations of r, picked from n distinct objects: n! / (r! (n-r)!).

The ! postfix means factorial.

Explanation

Let me explain using a very simple example: finding all combinations of 2 from a set of 6 letters {A, B, C, D, E, F}. The first combination is AB and the last is EF.

The total number of possible combinations is: n!/(r!(n-r)!)=6!/(2!(6-2)!)=15 combinations.

I'm thinking if you would have noticed by now, the number of times a letter appears. The formula for the number of times a letter appears in all possible combinations is n!/(r!(n-r)!) * r / n == c * r / n. Using the above example, it would be 15 * 4 / 6 = 10 times. All the letters {A, B, C, D, E, F} appear 10 times as shown. You can count them yourself to prove it.

Source Code Section

Please note that all the combination functions are now enclosed in the stdcomb namespace.

The Recursive Way

I have made a recursive function, char_combination() which, as its name implies, takes in character arrays and processes them. The source code and examples of using char_combination() are in char_comb_ex.cpp. I'll stop to mention that function. For now, our focus is on recursive_combination(), a template function, which I wrote using char_combination() as a guideline.

The parameters prefixed with 'n' are associated with the n sequence, while the r-prefixed one are r sequence related. As an end user, you need not bother about those parameters. What you need to know is func. func is a function defined by you. If the combination function finds combinations recursively, there must exist a way the user can process each combination. The solution is a function pointer which takes in two parameters of type RanIt (stands for Random Iterator). You are the one who defines this function. In this way, encapsulation is achieved. You need not know how recursive_combination() internally works, you just need to know that it calls func whenever there is a different combination, and you just need to define the func() function to process the combination. It must be noted that func() should not write to the two iterators passed to it.

The typical way of filling out the parameters is n_column and r_column is always 0, loop is the number of elements in the r sequence minus that of the n sequence, and func is the function pointer to your function (nbegin and nend, and rbegin and rend are self-explanatory; they are the first iterators and the one past the last iterators of the respective sequences).

Just for your information, the maximum depth of the recursion done is r+1. In the last recursion (r+1 recursion), each new combination is formed.

An example of using recursive_combination() with raw character arrays is shown below:

Certain conditions must be satisfied in order for next_combination() to work

All the objects in the n sequence must be distinct.

For next_combination(), the r sequence must be initialized to the first r-th elements of the n sequence in the first call. For example, to find combinations of r=4 out of n=6 {1,2,3,4,5,6}, the r sequence must be initialsed to {1,2,3,4} before the first call.

As for prev_combination(), the r sequence must be initialised to the last r-th elements of the n sequence in the first call. For example, to find combinations of r=4 out of n=6 {1,2,3,4,5,6}, the r sequence must be initialsed to {3,4,5,6} before the first call.

The n sequence must not change throughout the process of finding all the combinations, else results are wrong (makes sense, right?).

next_combination() and prev_combination() operate on data types with the == operator defined. That is to mean if you want to use next_combination() on sequences of objects instead of sequences of POD (Plain Old Data), the class from which these objects are instantiated must have an overloaded == operator defined, or you can use the predicate versions.

When the above conditions are not satisfied, results are undetermined even if next_combination() and prev_combination() may return true.

Return Value

When next_combination() returns false, no more next combinations can be found, and the r sequence remains unaltered. Same for prev_combination().

Some information about next_combination() and prev_combination()

The n and r sequences need not be sorted to use next_combination() or prev_combination().

next_combination() and prev_combination() do not use any static variables, so it is alright to find combinations of another sequence of a different data type, even when the current finding of combinations of the current sequence have not reached the last combination. In other words, no reset is needed for next_combination() and prev_combination().

Examples of how to use these two functions are in next_comb_ex.cpp and prev_comb_ex.cpp.

So what can we do with next_combination()?

With next_combination() and next_permutation() from STL algorithms, we can find permutations!!

The formula for the total number of permutations of the r sequence, picked from the n sequence, is: n!/(n-r)!

We can call next_combination() first, then next_permutation() iteratively; that way, we will find all the permutations. A typical way of using them is as follows:

However, I must mention that there exists a limitation for the above code. The n and r sequences must be sorted in ascending order in order for it to work. This is because next_permutation() will return false when it encounters a sequence in descending order. The solution to this problem for unsorted sequences is as follows:

However, this method requires you to calculate the number of permutations beforehand.

So how do I prove they are distinct permutations?

There is a set container class in STL we can use. All the objects in the set container are always in sorted order, and there are no duplicate objects. For our purpose, we will use this insert() member function:

pair <iterator, bool> insert(const value_type& _Val);

The insert() member function returns a pair, whose bool component returns true if an insertion is made, and false if the set already contains an element whose key had an equivalent value in the ordering, and whose iterator component returns the address where a new element is inserted or where the element is already located.

proof.cpp is written for this purpose, using the STL set container to prove that the permutations generated are unique. You can play around with this, but you should first calculate the number of permutations which would be generated. Too many permutations may take ages to complete (partly due to the working of the set container), or worse, you may run out of memory!

Thanks for your help and mail,sir. First I tried to run your example of char string i.e. n=123456 and r=1234 it runs fine, then I did what you said i.e. I did char [] as float [10] for n and r and then run in output it shows memory address for all pair (same value). Sir can you help me to solve this problem and second thing which I miss to mention in first mail is that I want to convert these code in ANSI C.
Thanking you.

Sir, Thanks for your help. Sir I just saw your part 2 upload, which is what I am looking for.I have implemented my algorithm in matlab which is serial now I want to convert it in c and parallel, and then test on cuda. so now I am understanding your "The Technique of Finding Combination, Given its Index" which you have explained very good, thanks. sir, I have not so far downloaded software and tested it, I will be very grateful to you if you help me to convert it in c language else I need to write my code in c++ which I don't know as I just started learning c language.

Sir, as you said in "A Benchmark Program for your PC" that "The program does not store the combination anywhere and discards them after every computation". If we want to see the combinations after computations, what and where we need to do the change in "TimeCombinations". Thanking you.
Sir, I also not find the source folder for "The Technique for Finding Combinations Without Repetitions" for example "Find Combinations of 3 from a Sequence of 5" with "Optimised Version: Index Combination". sir I will be very grateful if you can send/give idea where can I get it. Thanking you.

Thank you sir,in said folder I found the example by replacing the animal names by integers, it also works for floating numbers. sir I now started working on C++, as I found that more peoples are using it than C.

sir I repeat my first question again here,
Sir, as you said in "A Benchmark Program for your PC" that "The program does not store the combination anywhere and discards them after every computation". If we want to see the combinations after computations, what and where we need to do the change in "TimeCombinations". Thanking you.

sir, can you give more idea on how your code in "TimeCombinations" finds number of processors in cpu (as in my case 4) and split to find for combination for "91". Sir can it be possible if we use one thread in GPU for each combination. Thanking you.

Sir, there is correction in my previous reply, the correction is that in "TimeCombination" we don't find combination for 91, so my question is,on how your code in "TimeCombinations" finds number of processors in cpu (as in my case 4) and split to find on each processor. Sir can it be possible if we use one thread in GPU for each combination. Thanking you.